arXiv:1708.03789v1 [math.PR] 12 Aug 2017
The median for an exponential family and the normal law G´erard Letac∗, Mauro Piccioni†. August 15, 2017
Abstract Let P a probability on the real line generating a natural exponential family (Pt )t∈R . The property that t is a median of Pt for all t is fulfilled by the standard Gaussian law N (0.1). We show that this property is stable in the sense that if P has a bounded density with respect to N (0, 1) then P = N (0, 1). Keywords: Characterization of the normal laws, real exponential families, medians. MSC2010 classification: 62E10, 60E05, 42A38.
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Introduction
Let P be a probability on the real line and assume that for all t real Z L(t) = etx P (dx) < ∞.
(1)
R
Such a probability generates the natural exponential family FP = {Pt (dx) =
etx P (dx), t ∈ R}. L(t)
Suppose that the natural parameter t of FP is always a median of Pt , in the sense that for all t we have Pt ((−∞, t)) ≤
1 ≤ Pt ((−∞, t]). 2
(2)
In the sequel we denote by P the set of probabilities P such that (1) and (2) are fulfilled. The best example of an element of P is the standard normal distribution ∗ Laboratoire de Statistique et Probabilit´es, Universit´e Paul Sabatier, Toulouse, France.
[email protected], corresponding author. † Dipartimento di Matematica, Sapienza Universit`a di Roma, 00185 Rome, Italia.
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2 /2
N(0, 1) since in this case L(t) = et mention two simple facts about P.
and Pt = N(t, 1). The following propositions
Proposition 1. If P ∈ P then P has no atoms. As a consequence for all t Pt ((−∞, t]) = 21 . Proposition 2. If P ∈ P then for all −∞ ≤ a < b ≤ ∞ we have P ((a, b)) > 0. In other terms, the support of P is R and t is the unique median of Pt . One can formulate the following Conjecture 1. P = {N(0, 1)}. This aim of this note is to support Conjecture 1 by the following theorem. Theorem 3. Suppose that P ∈ P has the following form 1 2 dx P (dx) = f (x)e− 2 x √ 2π
where f is a bounded function. Then f = 1. Section 2 gives the proofs of the above results and Section 3 contains some remarks about Conjecture 1 and a companion Conjecture 2.
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Proofs
Proof of Proposition 1: Suppose that t0 is an atom of P . Choose ǫ > 0 and observe that Z ǫx L(t0 ) Pt0 +ǫ ((−∞, t0 + ǫ]) − Pt0 ((−∞, t0 + ǫ]) = e − 1 Pt0 (dx). L(t0 + ǫ) (−∞,t0 +ǫ] In a neighbourhood of 0 the integrand at the r.h.s. is bounded w.r.t. ǫ by an integrable function, hence by applying the dominated convergence theorem the integral converge to zero when ǫ → 0. As a consequence lim Pt0 +ǫ ((−∞, t0 + ǫ]) = Pt0 ((−∞, t0 ]) ≥ ǫ↓0
But since P ∈ P, from (2) we have Pt0 +ǫ ((−∞, t0 + ǫ]) ≤ Pt0 ((−∞, t0 ]) = 21 . In a similar way, one proves that
1 2
lim Pt0 +ǫ ((−∞, t0 − ǫ]) = Pt0 ((−∞, t0 )) ≤ ǫ↓0
1 . 2
, and this proves that 1 . 2
Since P ∈ P, from (2) we have Pt0 −ǫ ((−∞, t0 − ǫ]) ≥ 12 , and this proves that Pt0 ((−∞, t0 )) = 12 . As a result P ({t0 }) = 0, t0 cannot be an atom and the proposition is shown. 2
Proof of Proposition 2 If a = −∞ or b = +∞ then P and the Pt ’s are concentrated on a half line. So it is impossible to have a median in t for Pt if t is not in this half line. We now assume that a and b are finite. Denote Ft (x) = Pt ((−∞, x]). Since P ∈ P, from Proposition 1 we can claim that Ft (t) = 12 = 1 − Ft (t). However if a < t < b the hypothesis implies that Z a Z ∞ 1 1 1 tx e P (dx) = = 1 − Ft (b) = etx P (dx). Ft (a) = L(t) −∞ 2 L(t) b As a consequence for a < t < b we have Z a Z ∞ def def tx A(t) = e P (dx) = etx P (dx) = B(t) b
−∞
Now A and B are the Laplace transforms of the measures 1(−∞,a) (x)P (dx) and 1(b,∞) (x)P (dx) respectively, they are defined for all t in R and they are therefore the restriction to R of two entire functions which coincide on (a, b). As a consequence the two measures coincide, which is impossible. Proof of Theorem Recall that if g ∈ L1 (R) its Fourier transform is the function R 3 isx defined by gˆ(s) = R e g(x)dx. Let us rewrite the hypothesis as follows: for all t we have Z Z t 2 x2 1 1 tx− x2 √ f (x)dx = etx− 2 √ f (x)dx. 2 e 2π 2π R −∞ 2 We multiply both sides by e−t /2 : Z Z t 2 (t−x)2 1 1 − (t−x) e− 2 √ f (x)dx. 2 e 2 √ f (x)dx = 2π 2π R −∞ √ 2 In other terms, if k(x) = sign x e−x /2 / 2π the unknown function f satisfies Z k(t − x)f (x)dx = 0 (3) R
for all t ∈ R. To continue the proof we need a lemma.
ˆ Lemma 4: The Fourier transform k(s) is zero only for s = 0. Proof : For s > 0 we have r Z ∞ r Z ∞ x2 2 1 2 2 ˆ −ik(s) = e−y /2 sin(sy)dy = e− 2s2 sin x dx π s π 0 r 0 ∞ Z (k+1)π x2 1 2X e− 2s2 sin x dx = s π k=0 kπ ! r ∞ Z (2n+1)π Z (2n+2)π x2 x2 1 2X e− 2s2 sin x dx + = e− 2s2 sin x dx s π n=0 2nπ (2n+1)π r ∞ Z π (x+2nπ)2 (x+(2n+1)π)2 1 2X − − 2 2 2s 2s e −e sin x dx > 0. = s π n=0 0 3
Since k is an odd function then kˆ is odd as well. Since k is integrable then kˆ is ˆ continuous. Therefore k(s) = 0 if and only if s = 0. Next consider the functions x 7→ k(x − τ ), for all τ ∈ R, consider the linear space generated by these functions, namely by the functions of the form x 7→ h(x) =
n X i=1
ci k(x − τi )
(where n, (ci )ni=1 and (τi )ni=1 are arbitrary) and let Σ(k) be the closed linear span in R L1 (R) of this set of functions. In particular from (3) the equality R h(x)f (x)dx = 0 holds for any element h of Σ(k). Then recall the following Corollary 9.4 in section II from Korevaar (2004) (appearing in a more general form as Theorem IV in Wiener (1932)). ˆ Theorem 5. Let k be a function of L1 (R), such that k(s) 6= 0 for s 6= 0. Then any 1 L (R) function h such that Z ˆ h(0) = h(x)dx = 0 (4) R
belongs to Σ(k). With help of the previous results we can claim that all the functions hα,β , defined for any α < 0 < β hα,β (x) = 1/α for α < x < 0 = 1/β for 0 < x < β = 0 otherwise, are in Σ(k). Indeed it is immediately seen that Z ˆhα,β (0) = hα,β (x)dx = 0, which proves that (4) is satisfied. This implies that the relation Z hα,β (t − x)f (x)dx = 0 for all t ∈ R, holds for any α < 0 < β. By taking t = 0 one gets Z 0 Z Z 0 Z 1 1 1 β f (x)dx = f (x)dx f (x)dx = f (x)dx = |α| α β 0 −1 0 from which we can conclude that f (x) = c, for some c ∈ R. The proof is finished with the observation that f being a density w.r.t. the standard normal distribution, the constant c is constrained to be equal to 1. 4
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Concluding remarks
For dealing √ with2 Conjecture 1 in general one can introduce for P ∈ P the measure Q(dx) = 2πex /2 P (dx). Condition (1) implies that the unbounded measure Q must satisfy for all real t Z 1
e− 2 x
2 +xt
R
Q(dx) < ∞.
(5)
Conjecture 1 is equivalent to say that if Q satisfies (5) and the equation Z (k ∗ Q)(t) = k(t − x)Q(dx) = 0 R
for all t, then Q(dx) = dx. Theorem 3 solves this question under the extra hypothesis that Q(dx) = f (x)dx where f is bounded. It is also worthwhile to mention a companion conjecture about exponential families where median would coincide with the mean: Conjecture 2. Suppose that the probability P satisfies (1), and denote m(t) = R xPt (dx). If for all t real m(t) is a median of Pt , then P = N(m, σ 2 ) for some m R and σ. This conjecture leads to integral equations more complicate than k ∗ Q = 0, and the tools of harmonic analysis seem inappropriate here.
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References
Korevaar, J. (2004) Tauberian Theory, A Century of Developments, Grundleheren der mathematischen Wissenschaften 329, Springer-Verlag, Berlin. Wiener, N. (1932) ’Tauberian theorems’, Annals of Mathematics 33: 1-100.
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